Science:Math Exam Resources/Courses/MATH100/December 2019/Question 4 (b)/Solution 1

The Maclaurin polynomial is the same as the Taylor polynomial centered at 0. From inspection of T₂(x)=2 + 3 x² we know that $f(0)=2,\,f'(0)=0\,{\text{ and }}f''(0)=2!\cdot 3.$

On the other hand, the second Maclaurin polynomial for g(x) = e^x f(x) has the form $T_{2,g}(x)=g(0)+g'(0)x+{\frac {g''(0)}{2!}}x^{2}$.

Here, since e⁰=1, we have:

* g(0) = f(0)=2, 

* g'(x) = exf(x) + exf'(x), so that g'(0)= f(0) + f'(0)=2;

* g"(x) = exf(x) + 2exf'(x) + exf"(x) . This results in g"(0) = f(0) + 2f'(0) + f"(0)=8.

Putting everything together:

$T_{2,g}(x)=2+2x+4x^{2}$.